walid Posted November 25, 2006 Report Posted November 25, 2006 the following FM circuitI have tree questions:1- what the function of the 1 n cap2-why tapping the coil, why not connecting the antenna directly to the collector3-what the function of the 1.8p cap, and why it is so smallthanx Quote
audioguru Posted November 25, 2006 Report Posted November 25, 2006 the following FM circuitI have tree questions:1- what the function of the 1 n cap2-why tapping the coil, why not connecting the antenna directly to the collector3-what the function of the 1.8p cap, and why it is so smallHi Walid,That is an extremely simple FM transmitter circuit. Did you make it? How is its performance?1) The 1nF cap makes the transistor a common-base amplifier at radio frequencies where the emitter is its input and the collector is its output.2) The coil and capacitor in parallel with it is the tuned circuit that determines the RF frequency. The antenna is connected to it so anything that gets near or touches the antenna changes the capacitance and changes the tuned frequency. When the antenna is connected to a tap on the coil then the change of tuned frequency isn't as much, but then the range of the transmitter isn't as far.3) The 1.8pF cap provides positive feedback so the transistor oscillates. It is in parallel with the transistor's capacitance and the wiring capacitance so the total capacitance might be about 10pF. 10pF is not small at 100MHz where it has a reactance of 160 ohms. Quote
walid Posted November 25, 2006 Author Report Posted November 25, 2006 Thank you guru u are a good man1) The 1nF cap makes the transistor a common-base amplifier at radio frequencies where the emitter is its input and the collector is its output.why it need to be a common-base, what would be happen if we remove this 1n cap.I know that Quote
audioguru Posted November 25, 2006 Report Posted November 25, 2006 why it need to be a common-base, what would be happen if we remove this 1n cap?This circuit uses a common-base amplifier. There are many other kinds of oscillator circuits. The capacitor makes the base at 0V for RF frequencies. If it is removed then the circuit probably won't oscillate.I know that Quote
walid Posted November 25, 2006 Author Report Posted November 25, 2006 Hi gurui read carefull your last reply but it was not sufficient to make me understand so, i now reading this doc.http://www.maxim-ic.com/appnotes.cfm/appnote_number/2032may it help methen i'll back to u to finish this hard subjectthank u guru Quote
walid Posted November 25, 2006 Author Report Posted November 25, 2006 i read it but i not understand any thingit is not as i expectcan u guru point me to another article more close to what i needcan u help me in understanding carfully how this vco operatethank u Quote
audioguru Posted November 25, 2006 Report Posted November 25, 2006 Hi Walid,Maxim's VCO IC uses a varactor diode (it changes its capacitance when its voltage is changed) to frequency-modulate and to tune its carrier. Quote
walid Posted November 25, 2006 Author Report Posted November 25, 2006 Hi guruMaxim's VCO IC uses a varactor diode (it changes its capacitance when its voltage is changed) to frequency-modulate and to tune its carrier. ok, i agreemy circuit did not using a varactor diode, there is no varactor diode i think that a p-n junction do this, is this true and how? Quote
audioguru Posted November 25, 2006 Report Posted November 25, 2006 my circuit did not using a varactor diode, there is no varactor diode i think that a p-n junction do this, is this true and how?The datasheet for the transistor that is used in your project and for most other small transistors shows the transistor's change of its capacitance with voltage change, like a varactor diode.The input audio causes the transistor to conduct more and less which changes the voltage across it and changes its capacitance. Quote
walid Posted November 27, 2006 Author Report Posted November 27, 2006 Hi guruThis circuit uses a common-base amplifier. There are many other kinds of oscillator circuits. The capacitor makes the base at 0V for RF frequencies. If it is removed then the circuit probably won't oscillatewhy the circuit probably won't oscillate, can you please, explain this point more.The antenna impedance match to the circuit's output impedance is poor, but it is simple and it worksWhat would u do to improve the matching between o/p and antenna?The base input impedance of the transistor at low audio frequencies is its beta of about 200, times the emitter resistance of 220 plus its internal emitter resistance = 44.5k ohms. It is in parallel with the two 47k resistors also in parallel, so the load on the mic is 15.4k ohms. beta = 200, are u guess it or depend on some source or you consider any general purpose transistor's beta as approx 200.also you calculate the internal emitter resistance (re) as beta X re =500 ohm, that is re = 500/200 =2.5 ohm, frome this:25/Ic=2.5 ==> Ic = 10 mA, is these values from datasheets, I search for BC547 datacheets and find many copmanies, please provide me with your datasheet or tell me from any comany. It is complicated to calculate how much is the output voltage and current because tapping down the coil reduces the output voltage into the low impedance antenna but provides a better impedance match so the collector voltage swing is much higherBut with tap, the coil acts as a step down transformer?The input audio causes the transistor to conduct more and less which changes the voltage across it and changes its capacitance. Are you mean the capacitance between B and E?thank you alot Quote
audioguru Posted November 27, 2006 Report Posted November 27, 2006 why the circuit probably won't oscillate, can you please, explain this point more.The transistor needs a change of base to emitter current to be an amplifier and to oscillate. If the base voltage is not secured with a capacitor to ground then it will also have its voltage changed when the positive feedback changes the emitter voltage, then there is no change in base to emitter current. What would u do to improve the matching between o/p and antenna?I would add an RF amplifier that would have a lower output impedance than the oscillator. It would also isolate the antenna from the oscillator so the frequency wouldn't change when anything gets near the antenna. Then I would add a series or pi LC impedance-matching network at the output which would also reduce harmonics.also you calculate the internal emitter resistance (re) as beta X re =500 ohm, that is re = 500/200 =2.5 ohm, frome this:25/Ic=2.5 ==> Ic = 10 mA, is these values from datasheetsThe Re of a silicon transistor is approx. 26/Ie in mA, which is about 2.6 ohms. The beta ranges from 110 to 800 so I picked 200. You can check its accuracy by looking at the input impedance of a 2N3904 transistor on its datasheet since it is about the same as a BC547. At 10mA the input impedance of a typical 2N3904 is 500 ohms.But with tap, the coil acts as a step down transformer?Yes, it is an auto-transformer (single tapped coil).Are you mean the capacitance between B and E?The transistor's collector is connected to the tuned circuit. The collector to emitter capacitance changes which changes the tuned frequency at the collector. Quote
walid Posted November 28, 2006 Author Report Posted November 28, 2006 Hi guruThe transistor needs a change of base to emitter current to be an amplifier and to oscillate. If the base voltage is not secured with a capacitor to ground then it will also have its voltage changed when the positive feedback changes the emitter voltage, then there is no change in base to emitter current. I feel that i not understand what you want to say.Lets analyze that statement:I assume that VE = 3vVB = 3.7vpositive feedback changes the emitter voltage to say 4 vso VB = 4.7vnow what that cap do to maintain VB to 3.7 vthank u guru Quote
audioguru Posted November 28, 2006 Report Posted November 28, 2006 Hi Walid,A filter capacitor needs time to charge or discharge in order for its voltage to change. At 100MHz its value makes a good filter so its voltage doesn't change.At 7.5kHz and lower audio frequencies then the value of the capacitor makes a very poor filter so its voltage can change to allow audio to modulate the base of the transistor.In my FM transmitter I used a 470pF filter capacitor to allow modulation up to 15kHz. Quote
walid Posted December 10, 2006 Author Report Posted December 10, 2006 Hi guruThe base input impedance of the transistor at low audio frequencies is its beta of about 200, times the emitter resistance of 220 plus its internal emitter resistance = 44.5k ohms. It is in parallel with the two 47k resistors also in parallel, so the load on the mic is 15.4k ohms. This is good, but I want to introduce the effect of the 1nF cap on the lower 47Kohm.The reactance of that cap at freqs from 300Hz t0 15Khz is 10.6 Kohm to 53 Kohm47K // 10.6K = 8.65 K, and47K // 53K = 24.9Kso VB is not constant at VB = 9 * 47/(47+47) = 4.5v as i was thinkit is really changes from 1.4V to 3.12V with the audio freqNow Zin (total) = ZB//RB1//RB2since RB2 equivalant is change from 8.65 K to 24.9Kso Zin changeplease comment.It is complicated to calculate how much is the output voltage and current because tapping down the coil reduces the output voltage into the low impedance antenna but provides a better impedance match so the collector voltage swing is much higher. I agree that it is complicated to calculate how much is the output voltage and current, lets assume them and investigate the tapping effect on them:If the output voltage = 5mV rms at freq 100MHz as shown in the Fig. below, how much voltage reach the antenna?and what the value of the impedance the antenna see, if the coil is 330nH and its XL = 207 ohm at 100MHz?thank u guru Quote
audioguru Posted December 10, 2006 Report Posted December 10, 2006 Hi Walid,I started to calculate the effect of the 1nF capacitor on the frequency response of the electret mic at 15kHz but I made a simulation instead. It makes no difference! Quote
walid Posted December 10, 2006 Author Report Posted December 10, 2006 Hi guruFirst thank u for the figures from which i can see that the 1 nF has no resonable effect at AF.but u may not see this questionIt is complicated to calculate how much is the output voltage and current because tapping down the coil reduces the output voltage into the low impedance antenna but provides a better impedance match so the collector voltage swing is much higher. I agree that it is complicated to calculate how much is the output voltage and current, lets assume them and investigate the tapping effect on them:If the output voltage = 5mV rms at freq 100MHz as shown in the Fig. below, how much voltage reach the antenna?and what the value of the impedance the antenna see, if the coil is 330nH and its XL = 207 ohm at 100MHz?thank u guru these flowers are for you Quote
Kevin Weddle Posted December 10, 2006 Report Posted December 10, 2006 How do you know you have crossed the finish line with an oscillator? I seem to think that general test lab equipment cannot tell the whole story. I cannot believe these mass produced behemoth bluetooths with Intel Pentium 4's and Windows spyware can measure accurately enough for me.If you know the oscillator frequency, it's probably better to make your own measuring circuits. An embedded application. Quote
audioguru Posted December 10, 2006 Report Posted December 10, 2006 Hi Walid,Thanks for the white roses. They are fresh and smell nice. ;DThe tuned circuit at the collector of the transistor is resonant and tries to be a high impedance. The coil is an auto-transformer and the 50 ohm antenna is multiplied by the square of the turns ratio so appears as 5 x 5 x 50= 1250 ohms at the collector.I think the power output is higher than if the antenna is connected directly to the collector. Quote
walid Posted December 12, 2006 Author Report Posted December 12, 2006 Hi gurulets do it againat 106MHz, L =330nH and C=6.8pf, we haveXL = 2 pi f L = 220 ohmXC= 1/(2 pi f C) = 220 ohm (resonance)Ztot = (XL*XC)/(XL+XC) = 110 ohmI think it is wrong calc, Ztot must be zero because XL = j w Land XC = -1/j w cI don't know what to do............IF the LC impedance = 110 ohm as before, and the antenna impedance = 50 ohm110/50 = 2.2the turns ratio = sqr(2.2) = about 1.5:1 which equivalent to 3:2 = 6:4if the coil was 6 turns then the tap must be at the fourth turn and not the fivth turn to have better impedance matching.please guru correct me Quote
audioguru Posted December 12, 2006 Report Posted December 12, 2006 Hi Walid,A parallel resonant LC has a very high impedance, nearly infinite ohms. A series resonant LC has a very low impedance, the resistance of the coil plus the resistance of the wires on the capacitor. Look at LC Resonance in Google. Quote
walid Posted December 12, 2006 Author Report Posted December 12, 2006 Hi guru I go to google and search for LC Resonance and one of them was:http://www.pronine.ca/lcf.htmIt is a calculatorsubstitute L=330n and C=6.7pu will have f=106 MHz and Zo=220 ohm as i calculate it, it is not nearly infinite ohmsIt is very important to me to correctly calc the o/p Z of the oscillator because i want to discuss with u the pi C-L-C filter between this osc and the 50 ohm antenna from this site:http://hem.passagen.se/communication/antenimp.htmland to ask u why u don't use it in your FM Txdon't worry i understand 95% of what he said but still to know how to cal the o/p Z of the oscillator to match it with the antennaI make a page in MATHCAD for quickly calculationsthank u guru Quote
audioguru Posted December 12, 2006 Report Posted December 12, 2006 Hi Walid,The calculator you found is WRONG!It has the wrong formula for the impedance at resonance of a parallel LC.You shouldn't calculate things at such a high frequency because you don't know how much stray capacitance is in parallel with your C, and you don't know how much inductance is in the wiring of the L.I tested my FM transmitter only one day and didn't improve it by adding a matching network at its antenna. Its range and sound quality are much better than I expected and since my FM band is full, it causes interference. It is illegal (and not nice) to cause interference here.Here is another formula for the impedance at resonance of a parallel LC: Quote
walid Posted December 12, 2006 Author Report Posted December 12, 2006 Hi guruThe calculator you found is WRONG!It has the wrong formula for the impedance at resonance of a parallel LC.yes I agree, it It has the wrong formula and they calc it as 220 and i calc it the first time as 110 and mistake when tell you it 220 ( I was hurry)You shouldn't calculate things at such a high frequency because you don't know how much stray capacitance is in parallel with your C, and you don't know how much inductance is in the wiring of the L.yes, I know, but still must calc it to know the o/p impedance to correctly fit the matching network, then may modify it lettle to compensare for the stray capacitance and inductance Here is another formula for the impedance at resonance of a parallel LC: I calc it, using your formula, look below: Quote
audioguru Posted December 12, 2006 Report Posted December 12, 2006 No Walid.Your calculation doesn't include the +90 degrees and -90 degrees phase shifts of the capacitor and inductor and they cancel. Therefore at resonance a parallel LC has an infinitely high impedance. Quote
walid Posted December 13, 2006 Author Report Posted December 13, 2006 hi guruIf they cancel then the Z is zero not infinityI know it is not zero the ohmic resistance of the coil wire and cap materials .. etecplease look, the man who do this: http://hem.passagen.se/communication/antenimp.htmlsay: the matching between Zout and the 50 ohm antenna using Pi C-L-C filterif Zout is infinity, how can he put the matching networkto design the Pi C-L-C filter you must know Zout??Your last answers on this topic were not very goodit takes me up then down, I'm confused nowthank u Quote
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